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If an ideal gas is flowing with a velocity $v$, how is the equipartition theorem applied.

Normally, we can say that $\frac{1}{2}mv_{x,rms}^2=\frac{1}{2}k_BT$. We can do the same thing for $v_y$ &c. But, when a gas is flowing in the x direction, I don't think that $v_{x,rms}=v_{y,rms}$. I'm not too sure of this, the distribution may be such that the rms velocities are preserved. If the rms velocities aren't preserved, obviously we cannot use $\frac{1}{2}k_BT$ as the temperature is the same (or is it?). So how does one analyse such a situation with the equipartition theorem?

I'm not very good at Hamiltonian mechanics, so Wikipedia isn't helping.

Question sparked off by Air velocity in a double-skin facade

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up vote 1 down vote accepted

Do the work in the rest frame of the (bulk) gas. It's really that easy.

In other words for a gas whose bulk flow is along the $x$ axis the $y$ and $z$ velocity distributions are given by equipartition as is the deviation of the $x$ velocity from the bulk velocity.

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Makes sense; thanks! In retrospect it feels obvious.. All inertial frames are equivalent. –  Manishearth Feb 23 '12 at 5:22
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